Re: Is this a requirement for two groups to be isomorphic?

Originally Posted by x3bnm

For two groups to be isomorphic is it a requirement that binary operations in the two groups have to be different?Is there a situation where binary operations between two isomorphic groups are same? According to Mr. Pinter's Abstract Algebra on page-89 he defines isomorphism like this:
[I]Let and be groups. A bijective function with the property that for any two elements and in ,

I must wonder about the reason for your question. Not to be too pedantic, but Charles Pinter is not a good source for a text in algebra. He studied with major logicians in Paris and is known for his textbook in set theory.

I suggest that you refer to either Mac Lane or Herstein for definitions and examples of homomorphisms.

Re: Is this a requirement for two groups to be isomorphic?

Originally Posted by Plato

I must wonder about the reason for your question. Not to be too pedantic, but Charles Pinter is not a good source for a text in algebra. He studied with major logicians in Paris and is known for his textbook in set theory.

I suggest that you refer to either Mac Lane or Herstein for definitions and examples of homomorphisms.

Thanks Plato for recommendation. This book is just my first reading to get a sense of what abstract algebra is. Later I will definitely read what you recommended. Again thanks.

Re: Is this a requirement for two groups to be isomorphic?

For two groups to be isomorphic is it a requirement that binary operations in the two groups have to be different?

Is there a situation where binary operations between two isomorphic groups are same?

According to Mr. Pinter's Abstract Algebra on page-89 he defines isomorphism like this:

Let and be groups. A bijective function with the property that for any two elements and in ,

is called an isomorphism from to .

If there exists an isomorphism from to , we say that is isomorphic to .

Then in next page Mr. Pinter gives an example of two groups that have different binary operations but are isomorphic.

Does this mean(implicitly) that the binary operations in these two groups have to be different in order for them to be isomorphic? or am I wrong on this?

they can be, but they don't have to be. for example, one isomorphism is the following:

f:Z-->2Z, f(k) = 2k here not only is the operation (ordinary addition of integers) the same, we have the integers isomorphic to a proper subgroup! this is very strange, and fortunately can only happen with infinite groups.

but given an isomorphism f:G1-->G2 we can have any of the following situations:

G1 = G2, with the same operation (f does not need to be the identity, for example if G1 = G2 = Z, with the operation +, one isomorphism is given by f(x) = -x).
G1 = G2, with different operations (a common example is G1 = G2 = Z with the operation in G1 being ordinary addition, and * in G2 where x*y = x+y+1,

G1 ≠ G2, but have the same operation (as with my example above)
G1 ≠ G2, but have different operations (for example the group Z2xZ2 under addition modulo 2:

= {(0,0),(1,0),(0,1),(1,1)} (it helps to think of this as: "0 if even, 1 if odd") is isomorphic to the group:

where the operation is matrix multiplication.

however, isomorphisms are "too special" and they don't reveal enough information. it turns out that a more general kind of function between two groups is a HOMOMORPHISM:

f:G1-->G2, where f(a*1b) = f(a)*2f(b)

these are like isomorphisms, but they need not be bijective (in greek "iso-" means "equal" (from ισoς or "isos", "equal") and "homo-" means "similar" (from ὅμοιος or "homoios", "similar")), and are much more useful.

homomorphisms often confuse beginners, but i bet you've already seen this one:

Re: Is this a requirement for two groups to be isomorphic?

Deveno I understand what you said.
But why is it fortunate for a infinite group to be isomorphic to a proper subgroup but awkward if a finite group were isomorphic to a proper subgroup?

What would be wrong if a finite group were isomorphic to a proper subgroup?

I am sure that Deveno will answer. But I will point out that if two groups are isomorphic then they have the same cardinality. That can only happen between a group and a proper subgroup in infinite sets. It cannot happen with finite sets.

Re: Is this a requirement for two groups to be isomorphic?

"...we have the integers isomorphic to a proper subgroup! this is very strange, and fortunately can only happen with infinite groups..."

Sorry if I missed the point.

But why is it fortunate for a infinite group to be isomorphic to a proper subgroup but awkward if a finite group were isomorphic to a proper subgroup?

What would be wrong if a finite group were isomorphic to a proper subgroup?

Don't worry I found the answer. Because it's impossible for a [EDIT]finite[/EDIT] group of a set and another [EDIT]finite[/EDIT] group of it's proper subset be isomorphic. The word "proper" is the keyword. Thanks Deveno for all the help.

Re: Is this a requirement for two groups to be isomorphic?

For two groups to be isomorphic is it a requirement that binary operations in the two groups have to be different?

Is there a situation where binary operations between two isomorphic groups are same?

The reason "isomorphism" is important is that it is an "equivalence relation" which means, in particular, that it is "reflexive": any group is isomorphic to itself. So that, no, there is no necessity that the binary operations are different.

According to Mr. Pinter's Abstract Algebra on page-89 he defines isomorphism like this:

Let and be groups. A bijective function with the property that for any two elements and in ,

is called an isomorphism from to .

If there exists an isomorphism from to , we say that is isomorphic to .

Then in next page Mr. Pinter gives an example of two groups that have different binary operations but are isomorphic.

Does this mean(implicitly) that the binary operations in these two groups have to be different in order for them to be isomorphic? or am I wrong on this?

The fact the examples have different operations does not mean they MUST be different.